A series combination of `n_(1)` capacitors, each of value `C_(1)`, is charged by a source of potential difference `4 V`. When another parallel combination of `n_(2)` capacitors, each of value `C_(2)`, is charged by a source of potential difference `V`, it has same (total) energy stored in it, as the first combination has. the value of `C_(2)`, in terms of `C_(1)`, is then

Options

A. `(2C_(1))/(n_(1)n_(2))`
B. `(16C_(1))/(n_(1)n_(2))`
C. `(n_(1)n_(2))/(16C_(1))`
D. `2(n_(2))/(n_(1))C_(1)`

Answer

B

Explanation

When `n_(1)` capacitors each of capacity `C_(1)` are joined in series, theire resultant capacity is given by
`(1)/(C_(S))=(1)/(C_(1))+(1)/(C_(2))+ . . .+(1)/(C_(n))=(n)/(C)" "therefore C_(S)=(C_(1))/(n_(1))`
`therefore` The energy of the combination,
`(E_(1))=(1)/(2)CV_(1)^(2)=(1)/(2)[(C_(1))/(n_(1))](4V)^(2) ` . . . (1)
and when they are joined in parallel the resultant capacity,
`C_(P)=n_(2)C_(2)`
and the enegy of combination `E_(2)=(1)/(2)(n_(2)C_(2))V^(2)` . . (2)
`because E_(1)=E_(2)`
`therefore (1)/(2)((C_(1))/(n_(1)))16V^(2)=(1)/(2)n_(2)C_(2)V^(2)`
`therefore (16C_(1))/(n_(1))=C_(2)n_(2)" "therefore C_(2)=(16C_(1))/(n_(1)n_(2))`.